Q. An $\alpha$-particle moves in a circular path of radius $0.83\ cm$ in the presence of a magnetic field of $0.25\ Wb/m^{2}.$ The de Broglie wavelength associated with the particle will be
A) $1\ \mathring{\mathrm{A}}$
B) $0.1\ \mathring{\mathrm{A}}$
C) $10\ \mathring{\mathrm{A}}$
D) $0.01\ \mathring{\mathrm{A}}$
Explanation
(d) $\lambda = \frac{h}{p} \Rightarrow \lambda = \frac{h}{mv}$
$r = \frac{mv}{qB} \Rightarrow mv = qrB \Rightarrow (2e)\left( 0.83 \times 10^{- 2} \right)\left( \frac{1}{4} \right)$
$\lambda = \frac{6.6 \times 10^{- 34} \times 4}{2 \times 1.6 \times 10^{- 19} \times 0.83 \times 10^{- 12}} \Rightarrow \lambda = 0.01\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.